Past SAT Math, GAT / Qudurat, and AP Calculus AB weekly challenges collected for focused review, solution analysis, and smarter practice.
Each archived challenge was previously published through the StudyGlitch Weekly Math Challenge system. Use the archive to revisit solution paths, compare trap answers, and connect the result to diagnostic-style SAT, GAT, and AP math practice.
60 past weekly challenges found. Open one card at a time for the full solution review.
Solve a medium GAT quantitative fraction question about remaining water in a tank with full explanation.
Arithmetic and Fractions / Tracking remaining quantity after repeated fractional useA water tank was full. First, 25 of the water was used. Then 13 of the remaining water was used. If 12 liters were left, how many liters were in the tank when it was full?
B.
30After the first use, find the fraction remaining. Then take the second fraction from that remaining amount.
After using 25, the remaining water is:
1 - 25 = 35
Then 13 of the remaining water is used, so 23 of the remaining water is left.
23 × 35 = 25
So 25 of the tank equals 12 liters.
Full tank = 12 × 52 = 30
The second fraction is not taken from the original tank. It is taken from what remains after the first step. That is why the remaining fraction is 23 of 35, which equals 25 of the original tank.
A common mistake is subtracting 25 and 13 directly from the whole. The second fraction is not from the whole.
The trap is treating both fractions as parts of the original tank.
If this mistake happened, your issue may be base tracking in word problems, not fraction arithmetic itself.
In Qudurat questions, the wording “of the remaining” is a major signal.
Repeated fraction questions are usually about changing bases. Always ask: fraction of what?
Translate the story into remaining fractions instead of liters first.
Solve an SAT algebra free-response question involving a no-solution linear equation with parameters.
Algebra / Determining a parameter value that creates no solution in a linear equation5(kx − n) = −6514x − 8518
In the given equation, k and n are constants and n > 1. The equation has no solution. What is the value of k?
For a linear equation to have no solution, the coefficients of x must match but the constants must differ.
Expand the left side:
5(kx − n) = 5kx − 5n
The equation becomes:
5kx − 5n = −6514x − 8518
For no solution, the coefficient of x on both sides must be equal:
5k = −6514
k = −6570 = −1314
The condition n > 1 ensures the constant terms will not accidentally match. If −5n = −8518, then n = 1718, which is not greater than 1. So matching the x-coefficients gives the no-solution case.
A common mistake is trying to solve for n, but the question asks for k.
The trap is forgetting to divide by 5 after matching 5k to the right-side coefficient.
If this was difficult, it may show a weak spot in no-solution equation conditions.
This SAT algebra question tests parameter conditions in linear equations.
For no-solution linear equations, make the variable coefficients equal and constants unequal.
Compare coefficients after expanding the left side.
Practice an AP Calculus AB particle motion question estimating acceleration from a velocity table.
Applications of Differential Calculus / Estimating acceleration as the derivative of velocityThe table shows the velocity at various times of an object moving along a line. An estimate of its acceleration, in ft/sec2, at t=1 is
| t (sec) | 1.0 | 1.5 | 2.2 | 2.5 |
|---|---|---|---|---|
| v (ft/sec) | 12.2 | 13.0 | 13.4 | 13.7 |
D.
1.6Acceleration is the rate of change of velocity.
Estimate acceleration at t=1 using the nearest available interval, from t=1.0 to t=1.5.
a(1)≈13.0−12.21.5−1.0=0.80.5=1.6
Acceleration is the derivative of velocity. With tabular data, a nearby secant slope estimates the derivative.
A common mistake is choosing 0.8, which is the change in velocity but not the rate of change per second.
The trap is forgetting to divide by the time difference.
If this was difficult, it may reveal weakness in interpreting acceleration as dv/dt.
This AP Calculus AB applications question tests derivative estimation from a velocity table.
Use slope of velocity versus time to estimate acceleration.
Use the closest time interval to t=1.
Practice a hard GAT quantitative chained-ratio question by matching the shared term.
Ratios and Proportions / Combining two ratios that share a common quantityIf a:b = 3:4 and b:c = 6:5, what is the ratio a:c?
A.
9:10Make the value of b the same in both ratios.
The ratios are:
a:b = 3:4
b:c = 6:5
The shared term is b. Make the b-values equal. The least common multiple of 4 and 6 is 12.
Multiply a:b = 3:4 by 3:
a:b = 9:12
Multiply b:c = 6:5 by 2:
b:c = 12:10
So:
a:c = 9:10
Because b appears in both ratios, it must represent the same amount in both. After matching b to 12, the corresponding values are a = 9 and c = 10.
A common mistake is combining 3 directly with 5 and choosing 3:5. The middle term b must be matched first.
The trap is ignoring that b has two different ratio values before adjustment.
If this was missed, the issue may be linking ratios through a shared variable.
GAT chained-ratio questions test whether you control the common term.
For chained ratios, equalize the shared quantity before comparing the outside quantities.
LCM of the two b parts: 4 and 6 gives 12.
Solve an SAT advanced algebra free-response question using the discriminant to find an integer parameter.
Advanced Algebra / Using the discriminant to find an integer parameter for no real solutionsr2 + qr = 8r − 87
In the given equation, q is an integer constant. The given equation has no real solutions. What is the largest possible value of q?
Move all terms to one side and use the discriminant condition for no real solutions.
Rewrite the equation:
r2 + qr = 8r − 87
r2 + (q − 8)r + 87 = 0
For no real solutions, the discriminant must be negative:
(q − 8)2 − 4(1)(87) < 0
(q − 8)2 < 348
Since √348 is between 18 and 19, the greatest integer value of q occurs when:
q − 8 = 18
q = 26
The discriminant must be strictly less than zero. Since (q−8)2 must be less than 348, the greatest integer value for q−8 is 18, giving q=26.
A common mistake is rounding √348 up to 19 and choosing 27. But 192 = 361, which is too large.
The trap is using ≤ or rounding the square root incorrectly.
If this was difficult, it may show a weak spot in discriminant inequalities.
This SAT advanced algebra question tests quadratic discriminants and integer constraints.
For no-real-solution questions, use discriminant less than zero.
Since 182 = 324 and 192 = 361, use 18.
Practice an AP Calculus AB definite integral question using the Fundamental Theorem of Calculus and table values.
Definite Integrals / Using the Fundamental Theorem of Calculus with table valuesThe table above shows some values of continuous function f and its first derivative. Evaluate ∫80 f′(x) dx.
| x | f(x) | f′(x) |
|---|---|---|
| 0 | 11 | 3 |
| 2 | 15 | 2 |
| 4 | 16 | −1 |
| 6 | 12 | −3 |
| 8 | 7 | 0 |
C.
4Use the Fundamental Theorem of Calculus: ∫ f′(x)dx gives change in f.
By the Fundamental Theorem of Calculus,
∫80 f′(x)dx = f(0) − f(8)
From the table, f(0)=11 and f(8)=7. Therefore,
f(0) − f(8) = 11 − 7 = 4
The derivative values in the table are not needed here. The integral of f′ over an interval gives the net change in f.
A common mistake is trying to approximate the integral from the derivative values instead of using f(0) and f(8).
The trap is ignoring the order of the bounds.
If this was difficult, it may reveal weakness in FTC endpoint logic.
This AP Calculus AB definite integral question tests the Fundamental Theorem of Calculus.
When integrating a derivative, go directly to endpoint values of the original function.
The reversed bounds mean f(0)−f(8), not f(8)−f(0).
Practice a medium GAT geometry question using the exterior angle theorem in a triangle.
Geometry / Using exterior angle relationships in trianglesIn a triangle, two interior angles are 48° and 67°. What is the measure of the exterior angle at the third vertex?
D.
115°An exterior angle equals the sum of the two remote interior angles.
The exterior angle at the third vertex equals the sum of the other two interior angles:
48° + 67° = 115°
So the exterior angle is 115°.
You could also find the third interior angle first: 180° - 48° - 67° = 65°. The exterior angle adjacent to it is 180° - 65° = 115°. Both methods give the same answer.
A common mistake is choosing 65°, which is the third interior angle, not the exterior angle.
The trap is finding the missing interior angle and stopping too early.
If this was missed, the issue may be angle-label attention under time pressure.
GAT geometry questions often test whether you answer the requested angle, not just the first angle you find.
For triangle exterior angle questions, either add the two remote interior angles or subtract the adjacent interior angle from 180°.
Use the exterior angle shortcut: remote angle + remote angle.
Solve an SAT statistics question involving frequency tables, weighted means, and a negative constant.
Statistics and Data Analysis / Calculating means from frequency tables with algebraic valuesThe frequency tables represent data sets A and B, where c is a negative integer constant. The mean of data set A is r and the mean of data set B is q. What is the value of rq?
Data Set A
| Value | Frequency |
|---|---|
| c | 12 |
| 2c | 21 |
| 3c | 30 |
Data Set B
| Value | Frequency |
|---|---|
| c | 30 |
| 2c | 21 |
| 3c | 12 |
C.
43Compute each weighted mean. The value of c will cancel in the ratio.
For data set A, the total frequency is:
12 + 21 + 30 = 63
The weighted sum is:
12c + 21(2c) + 30(3c) = 12c + 42c + 90c = 144c
So:
r = 144c63 = 16c7
For data set B, the weighted sum is:
30c + 21(2c) + 12(3c) = 30c + 42c + 36c = 108c
So:
q = 108c63 = 12c7
Therefore:
rq = 16c712c7 = 43
The fact that c is negative does not change the ratio because c appears in both means and cancels. The two means are different because the frequencies for c and 3c are reversed.
A common mistake is thinking the ratio cannot be determined because c is unknown. Since both means are multiples of c, it cancels.
The trap is treating the unknown negative constant as preventing calculation.
If this was difficult, it may show a weak spot in weighted averages with symbolic values.
This SAT statistics and data analysis question tests weighted means and algebraic cancellation.
For frequency-table mean questions, multiply each value by its frequency before dividing by total frequency.
The total frequency is the same for both data sets, which makes the ratio easier to compare.
Practice an AP Calculus AB differentiation question using the inverse function derivative formula.
Differentiation / Using the inverse function derivative formulaLet f be differentiable and one-to-one. If
f(3)=8
and
f′(3)=4
then
(f−1)′(8)
is
A.
14Use the input that maps to 8.
The inverse derivative formula is (f−1)′(a)=1f′(f−1(a)). Since f(3)=8, f−1(8)=3. Therefore, (f−1)′(8)=1f′(3)=14.
The derivative of an inverse function at an output value uses the reciprocal of the original derivative at the corresponding input value.
A common mistake is using f′(8), which is not given and is not the correct input.
The trap is confusing the inverse input with the original input.
If this was difficult, it may reveal weakness in inverse derivative notation.
This AP Calculus AB differentiation question tests inverse function derivatives.
For inverse derivatives, first find the original input associated with the given inverse input.
Since f(3)=8, the relevant original input is 3.
Solve a medium GAT quantitative age word problem involving age difference and future total.
Word Problems and Logic / Translating age relationships into equationsAhmed is 4 years older than Sami. In 6 years, the sum of their ages will be 50. How old is Ahmed now?
B.
21Let Sami’s current age be x. Then Ahmed’s current age is x + 4.
Let Sami’s current age be x. Then Ahmed’s current age is x + 4.
In 6 years:
Sami = x + 6
Ahmed = x + 10
Their sum will be 50:
(x + 6) + (x + 10) = 50
2x + 16 = 50
2x = 34
x = 17
Ahmed is now:
17 + 4 = 21
The future sum includes both people aging by 6 years, so the total increases by 12 years. The current total is therefore 50 - 12 = 38. Since Ahmed is 4 years older, Ahmed is 21 and Sami is 17.
A common mistake is adding 6 years only once. Both people age by 6 years.
The trap is applying the future change to only one person.
If this was missed, the issue may be translating time-shifted word problems into equations.
GAT age questions often test whether you apply time changes to every person involved.
For age problems, track each person separately when moving into the future.
Since two people age by 6 years each, the total age increases by 12.
Solve an SAT advanced algebra question about the coefficient of x to the fifteenth in a polynomial product.
Advanced Algebra / Finding a specific coefficient in a product of polynomialsf(x) = 11x10 + 3x8
g(x) = −17x7 + 9x5
The polynomial p(x) is defined as the product of the given polynomials, f(x) and g(x). What is the coefficient of x15 in p(x)?
B.
48Only multiply term pairs whose exponents add to 15.
The x15 terms come from:
(11x10)(9x5) = 99x15
and:
(3x8)(−17x7) = −51x15
Add the coefficients:
99 + (−51) = 48
You do not need to expand the entire product. Focus only on the term pairs that create x15.
A common mistake is using only one contributing pair and choosing 99.
The trap is forgetting the negative contribution from 3x8 · −17x7.
If this was difficult, it may show a weak spot in polynomial products and coefficient tracking.
This SAT advanced algebra question tests polynomial multiplication and targeted coefficient extraction.
For coefficient questions, identify all exponent pairs that produce the target power.
Look for exponent sums of 15: 10+5 and 8+7.
Practice an AP Calculus AB functions question about zeros of a composite function.
Functions / Finding zeros of a composite functionLet
g(x)=f(x2−1)
If f(a)=0 only when a=3, then the zeros of g are
C.
x=−2 and x=2Set the input to f equal to the value that makes f zero.
For g(x)=0, we need f(x2−1)=0. Since f(a)=0 only when a=3, set x2−1=3. Then x2=4, so x=−2 or x=2.
The zeros of the composite function happen when the inside expression produces the zero-input of the outside function.
A common mistake is setting x2−1=0 instead of setting it equal to 3.
The trap is solving for zeros of the inside expression instead of zeros of the composite.
If this was difficult, it may reveal weakness in composite-function reasoning.
This AP Calculus AB functions question checks composite-function structure.
For composite-function zeros, identify the required input to the outer function.
The input x2−1 must equal 3.